Question

A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 300 μC. When potential across the capacitor is reduced by 100 V, the charge stored in it becomes 100 μC. Calculate The potential V and the unknown capacitance. What will be the charge stored in the capacitor if the voltage applied had increased by 100 V?

Answer 1

(i) Initial voltage, V₁ = V volts, Charge stored,Q₁ = 300 µC

Q₁ = CV₁ ..(1)

Changed potential, V₂ = V-100 V

Q₂ = 100 µC

Q₂ = CV₂ ..(2)

By dividing (2) from (1), we get `(Q_1)/(Q_2) =  (CV_1)/(CV_2) => Q_1/Q_2 = V_1 /V_2`

`=> 300/100 = V/(V- 100) => V = 150 \text { volts }`

`∴ C = Q_1/V_1 = (300 xx 10^-6)/150  = 2 xx 10^-6 F = 2 μ F`

 

(ii) If the voltage applied had increased by 120 V, then V₃ = 150+100 = 250 V

Hence, charge stored in the capacitor, Q₃ = CV₃ = 2×10^-6×250 = 500 µC

(ii) If the voltage applied had increased by 120 V, then V₃ = 150+100 = 250 V

Hence, charge stored in the capacitor, Q₃ = CV₃ = 2×10^-6×250 = 500 µC